How do you find the domain and range of $x/(x^2+1)$ ?

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Answer 1 · 2015-09-18T21:13:21

The domain is $R$ and the range is $[-1/2,1/2]$

The domain is $R$ the set of all reals because $1+x^2>0$ .

To find the range we have that

$y=x/(x^2+1)=>y(x^2+1)=x=>yx^2+x+y=0=$

this is trinomial with respect to x hence

$1^2-4y*y>=0=>1-4y^2>=0=>-1/2<=y<=1/2$

How do you find the domain and range of x/(x^2+1)x/(x^2+1)?